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Forgive me for placing extra-large image right next to a tiny one; I don't know how to change it.

Equating forces on the sphere along the perpendicular of the surface of the wedge , $N(1+\sin^2 \theta) = mg\cos \theta + f\cos \theta\sin\theta\\\Rightarrow \dfrac {34N}{25}= \dfrac{12f}{25} + \dfrac{4mg}5~~\cdots(1)$

Also, $\alpha = \dfrac{5f}{2mR}\\ma = mR\alpha = \dfrac{5f}2$

Taking forces along the surface of the wedge, $N\sin\theta\cos\theta + mg\sin\theta - f(1+\cos^2\theta) = \dfrac{5f}2~~\cdots(2)$

Substituting $(1)$ in $(2)$ , $144f+ 48\times5mg+510 mg=3519f\\f = \dfrac{750}{3375}mg=\dfrac 29 mg$

$\Huge\therefore \boxed{f=\dfrac29mg}$

$thatas how i didit$