Answer
12
Work Step by Step
At the 95% confidence, the critical z-value is $z_{\alpha/2}=1.96$
Given $\sigma=0.26,E=0.15$, the margin of error can be
expressed as $E=1.96\times\frac{0.26}{\sqrt {n}}=0.15$ which gives
the sample size as $n=(\frac{1.96\times0.26}{0.15})^2\approx12$ (round up to the next integer here)