First, to define the 'safe distance'. The most logical safe distance would be defined as being just outside the event horizon of the black hole which is the distance from the black hole at which not even light can escape. Sub-light speed orbits can still be stable outside the event horizon.
For a black hole, the event horizon's physical extent is defined as the Schwarzschild radius, 
$r_s$, given by the equation:
$$r_s = \frac{2GM}{c^2}$$
where M is the black hole mass, c is the speed of light, and G is the gravitational constant.
Therefore, given the above equation, the angular extent (in radians) of the black hole's event horizon is given by:
$$\theta = 2tan^{-1}(\frac{GM}{c^2d})$$
where the factor 2 comes from moving from radius to diameter.
Since you have fixed the mass of the black hole to 
$10^8M_\odot$, its angular extent is purely a function of observer distance from the black hole, as plotted below (in units of degrees):
Therefore, with the black hole centred on our sun, the event horizon would be just inside the Earth's orbit. Meaning if the Earth was in the correct orbit, you'd be safe to observe the black hole. To put the angular extent further into context, if we were sat on Pluto at the farthest point in its orbit, the event horizon would be over 4 times larger than the full moon as seen from Earth!
So I'd say, given the apparent planetary orbits of Interstellar's exoplanetary system, Gargantua's extent is pretty realistic... Why there'd be a stable planetary system in orbit about a supermassive black hole in the first place is probably the more unbelievable bit!
PS - This answer disregards the accretion disc seen around the black hole and only concentrates in the angular extent of the event horizon.
