$P(n = even) = \frac{5}{6}\cdot \frac{1}{6} + \frac{5}{6}\cdot \left (\frac{5}{6} \right )^{2} \cdot \frac{1}{6} + \frac{5}{6}\cdot \left (\frac{5}{6} \right )^{4} \cdot \frac{1}{6} +\frac{5}{6}\cdot \left (\frac{5}{6} \right )^{6} \cdot \frac{1}{6} + ... $
$\implies P(n=even) = \frac{\frac{5}{6}\cdot \frac{1}{6}}{1 – \left (\frac{5}{6} \right )^{2} } = \frac{5}{11} $
$P({(n=4)}\mid{ (n=even)}) = \frac{ \frac{5}{6}\cdot \left (\frac{5}{6} \right )^{2} \cdot \frac{1}{6}}{\frac{5}{11}} = \frac{5^3\cdot 11}{5\cdot 6^4} = \frac{5^2\cdot 11}{6^4}$
Hence D) none of the above.