An inspiration

Consider:

$\dfrac AB = \dfrac {202^{303}}{303^{202}} = \dfrac {2^{303}\times 101^{303}}{3^{202}\times 101^{202}} = \dfrac {8^{101}\times 101^{101}}{9^{101}} = \left(\dfrac {808}9 \right)^{101} > 1$

$\implies \boxed{A>B}$

$A = \small202^{303} = (2*101)^{303} = (2)^{303}*(101)^{303}$

$B = \small303^{202} = (3*101)^{202} = 3^{202} *(101)^{202}$ $\small\frac{A}{B} = \left(\frac{2^{303}}{3^{202}}\right)*\left(\frac{101}{101}\right)^{303-202} = \left(\frac{8×101}{9}\right)^{101} = \left(\frac{808}{9}\right)^{101} > 1$ Hence, $\boxed{A>B}$