Confidence Intervals · Superchallenge problem

A pollster does not trust the normal approximation and instead uses Chebyshev's inequality, $P(|\hat p - p| \ge E) \le \dfrac{\operatorname{Var}(\hat p)}{E^2}$, together with the worst case $\operatorname{Var}(\hat p) \le \dfrac{1}{4n}$. What is the smallest sample size $n$ that GUARANTEES $P(|\hat p - p| < 0.05) \ge 0.95$ by this argument?