First, for simplicity, I am assuming that the initial substrate (2-iodobutane with radioactive iodine) consists of only one enantiomer.

how would 2-iodo butane lose its optical activity(it still has a chiral C).

In $SN_1$ mechanism, there is 50% inversion and 50% retention, so a racemic mixture starts to form which decreases the optical activity of reaction mixture due to external compensation.

In easy terms: In a racemic mixture, although the two enantiomers rotate plane-polarized light in opposite directions, the rotations cancel because they are present in equal amounts. And in a partially racemic mixture or still undergoing reaction mixture with racemisation, the optical activity is decreased (rotation of plane polarised light is less).

So, 2-iodobutane doesn’t lose its optical activity, it’s optical activity is only countered by its inverted enantiomer.

What percentage of reaction proceeds via SN1 mechanism?

The rate of loss of radioactivity is proportional to amount of substrate that has reacted (2-iodobutane with radioactive iodine), that is rate of reaction.

And, the rate of loss of optical activity is proportional to rate formation of racemic mixture.

Now suppose,

$x =$ rate of reaction of substrate at an instant in moles per unit time.

Then, $1.96x$ is the rate of racemisation.

So, during a small time period $t$, $xt$ moles of substrate has reacted and $1.96xt$ moles of racemic mixture is obtained.

The racemic mixture consists of 50% inverted product and 50% with original configuration. Note that the reaction mixture is not purely racemic. It has substrate and retention product (both of which have original configuration) and inversion product. So, by racemic mixture I mean all the inverted product and equal amount of molecules with original configuration.

Since no inverted molecule was present before the reaction (assumption at start of answer), $0.98xt$ moles of substrate proceeded via inversion and the rest $0.02xt$ moles via retention ($x$ moles had reacted)

So, only $SN_1$ can cause retention and that also only 50% of reaction via $SN_1$ gives retention product, $0.04xt$ moles of substrate must have proceeded via $SN_1$ mechanism.

Thus, percentage of $SN_1$ mechanism = $\frac{0.04x}{x}$ x $100$% $=4$%

First, for simplicity, I am assuming that the initial substrate (2-iodobutane with radioactive iodine) consists of only one enantiomer.

How would 2-iodobutane lose its optical activity  (it still has a chiral $\ce{C}$)?

In $\mathrm{S_N1}$ mechanism, there is 50% inversion and 50% retention, so a racemic mixture starts to form which decreases the optical activity of reaction mixture due to external compensation.

In easy terms: In a racemic mixture, although the two enantiomers rotate plane-polarized light in opposite directions, the rotations cancel because they are present in equal amounts. And in a partially racemic mixture or still undergoing reaction mixture with racemisation, the optical activity is decreased (rotation of plane polarised light is less).

So, 2-iodobutane doesn’t lose its optical activity, it’s optical activity is only countered by its inverted enantiomer.

What percentage of reaction proceeds via $\mathrm{S_N1}$ mechanism?

The rate of loss of radioactivity is proportional to amount of substrate that has reacted (2-iodobutane with radioactive iodine), that is rate of reaction.

And, the rate of loss of optical activity is proportional to rate formation of racemic mixture.

Now suppose,

$x =$ rate of reaction of substrate at an instant in moles per unit time.

Then, $1.96x$ is the rate of racemisation.

So, during a small time period $t$, $xt$ moles of substrate has reacted and $1.96xt$ moles of racemic mixture is obtained.

The racemic mixture consists of 50% inverted product and 50% with original configuration. Note that the reaction mixture is not purely racemic. It has substrate and retention product (both of which have original configuration) and inversion product. So, by racemic mixture I mean all the inverted product and equal amount of molecules with original configuration.

Since no inverted molecule was present before the reaction (assumption at start of answer), $0.98xt$ moles of substrate proceeded via inversion and the rest $0.02xt$ moles via retention ($x$ moles had reacted)

So, only $\mathrm{S_N1}$ can cause retention and that also only 50% of reaction via $\mathrm{S_N1}$ gives retention product, $0.04xt$ moles of substrate must have proceeded via $\mathrm{S_N1}$ mechanism.

Thus, percentage of $\mathrm{S_N1}$ mechanism $= \dfrac{0.04x}{x} \times 100\% =4\%$.