a
(a) Angle of repose \(\alpha = {\tan ^{ - 1}}(\mu ) = {\tan ^{ - 1}}(0.8) = 38.6^\circ \)

Angle of inclined plane is given \(\theta = 30^\circ \).

It means block is at rest therefore,
Static friction \(=\) component of weight in downward direction \( = mg\sin \theta = 10\;N\)

\(\therefore m = \frac{{10}}{{9 \times \sin 30^\circ }} = 2\;kg\)