From the above figure,

\(Mg \sin \theta- f = ma _{ cm }\)

As we know that,

\(\tau_{ cm }= I _{ cm } \alpha\)

And,

\(f =\frac{ MR ^{2}}{2}\left(\frac{ a _{ cm }}{ R ^{2}}\right)\)

\(=\frac{ Ma _{ cm }}{2}\)

\(a_{c m}=\frac{2 f_{R}}{M}\)

Therefore,

\(mg \sin \theta- f = M \left(\frac{2 f }{ M }\right)\)

\(f =\frac{ mg \sin \theta}{3}\)

Substitute the values.

\(f =\frac{1}{2}\left(\frac{10}{3}\right)\left(\frac{1}{\sqrt{2}}\right)\)

\(=\frac{5}{3 \sqrt{2}} N\)