Given,

$${{{r_A}} \over {{r_B}}} = {2 \over 3}$$

$${{{\rho _A}} \over {{\rho _B}}} = {3 \over 5}$$

We know,

Acceleration due to gravity

$$g = {{GM} \over {{r^2}}} = {G \over {{r^2}}} \times {4 \over 3}\pi {r^3} \times \rho $$

$$ = {{4\pi Gr\rho } \over 3}$$

$$\therefore$$ $$g \propto \rho r$$

$$\therefore$$ $${{{g_A}} \over {{g_B}}} = {{{\rho _A}} \over {{\rho _B}}} \times {{{r_A}} \over {{r_B}}}$$

$$ = {3 \over 5} \times {2 \over 3}$$

$$ = {2 \over 5}$$