We have $a^b<b^a$ iff $b\log(a)<a\log(b)$ iff $\frac{\log(a)}a<\frac{\log(b)}b$. Consider the function $f:x\mapsto\frac{\log(x)}x:\mathbb R^+\to\mathbb R$. Then $f'(x)=\frac{1-\log(x)}{x^2}$, hence $f'(x)>0$ iff $x<e$, hence $f$ is decreasing for $x>e$ and this proves $f(303)<f(202)$, hence $303^{202}<202^{303}$.

We have $a^b<b^a$ iff $b\log(a)<a\log(b)$ iff $\frac{\log(a)}a<\frac{\log(b)}b$. Consider the function: $$f:x\mapsto\frac{\log(x)}x:\mathbb R^+\to\mathbb R$$ Then: $$f'(x)=\frac{1-\log(x)}{x^2}$$ Hence $$f'(x)>0\quad\mathrm{iff}\quad x<e$$ Hence $f$ is decreasing for $x>e$ and this proves $f(303)<f(202)$, hence $303^{202}<202^{303}$.