Answer
The tension in the cable is $~~76~N$
Work Step by Step
Let $F_N$ be the normal force on the sphere from the plane.
Let $F_T$ be the tension in the cable.
The horizontal component of $F_N$ is equal in magnitude to the horizontal component of $F_T$:
$F_N~sin~\theta = F_T~cos~(\theta-\phi)$
$F_N = \frac{F_T~cos~(\theta-\phi)}{sin~\theta}$
The sum of the vertical components of $F_N$ and $F_T$ is equal in magnitude to the weight of the sphere:
$F_N~cos~\theta+F_T~sin~(\theta-\phi) = mg$
$\frac{F_T~cos~(\theta-\phi)}{sin~\theta}~cos~\theta+F_T~sin~(\theta-\phi) = mg$
$F_T~cos~(\theta-\phi)~cot~\theta+F_T~sin~(\theta-\phi) = mg$
$F_T = \frac{mg}{cos~(\theta-\phi)~cot~\theta+sin~(\theta-\phi)}$
$F_T = \frac{(10~kg)(9.8~m/s^2)}{cos~(45^{\circ}-25^{\circ})~cot~45^{\circ}+sin~(45^{\circ}-25^{\circ})}$
$F_T = 76~N$
The tension in the cable is $~~76~N$
