Answer
$6\frac{1}{4}\%$ compounded annually
Work Step by Step
Case 1. Given $r=0.06$, let $t=1$, we have $A_1=P(1+\frac{0.06}{4})^{4}\approx1.0614P$
Case 2. Given $r=0.0625$, let $t=1$, we have $A_2=P(1+\frac{0.0625}{1})^{1}\approx1.0625P$
Thus case 2 with $6\frac{1}{4}\%$ compounded annually is better.