Given
Initial conditions:
$P_1 = 74 cm$
$T_1 = 20^{\large\circ}C = (20 + 273) = 293 K$
$V_1 = 512 mL$
Final conditions (STP):
$P_2 = 76 cm$
$T_2 = 0^{\large\circ}C = 273 K$
$V_2 = ?$
We know that
$\large\frac{P_1V_1}{T_1} = \large\frac{P_2V_2}{T_2}$
$\large\frac{74\times512}{293} = \large\frac{76\times V_2}{273}$
$V_2 = \large\frac{74\times512\times273}{293\times76}$
= 464.5 mL
Hence answer is (B)