Molarity of original solution (M1) = 3 M
Volume of original solution ($V_1$) = ?
Molarity of final solution ($M_2$) = 2 M
Volume of final solution ($V_2$) = 0.060 L = 60 mL
By law of dilution, $V_1 M_1 = V_2M_2$
$V_1=\large\frac{2\times 60}{3}$$=40mL$
Volume of water added = 60 mL − 40 mL = 20 mL
Hence (A) is the correct answer.