Answer
The tension in the top wire is 14.2 N
The tension in the bottom wire is 8.3 N
Work Step by Step
We can find the angle $\theta$ that each wire makes with the vertical.
$cos(\theta) = \frac{0.5~m}{1.0~m}$
$\theta = arccos(\frac{0.5~m}{1.0~m})$
$\theta = 60^{\circ}$
We can find the radius of the circle that the ball rotates.
$\frac{r}{1.0~m} = sin(60^{\circ})$
$r = 0.866~m$
Let $T_t$ be the tension in the top wire. Let $T_b$ be the tension in the bottom wire. The vertical component of $T_t$ is equal to the sum of the vertical component of $T_b$ and the ball's weight.
$T_t~cos(\theta) = T_b~cos(\theta) + mg$
$T_t = \frac{T_b~cos(\theta) + mg}{cos(\theta)}$
The sum of the horizontal component in each wire provides the centripetal force to keep the ball moving in a circle.
$T_t~sin(\theta)+T_b~sin(\theta) = \frac{mv^2}{r}$
$(\frac{T_b~cos(\theta) + mg}{cos(\theta)})~sin(\theta)+T_b~sin(\theta) = \frac{mv^2}{r}$
$mg~tan(\theta)+2~T_b~sin(\theta) = \frac{mv^2}{r}$
$T_b = \frac{\frac{mv^2}{r}-mg~tan(\theta)}{2~sin(\theta)}$
$T_b = \frac{\frac{(0.300~kg)(7.5~m/s)^2}{0.866~m}-(0.300~kg)(9.80~m/s^2)~tan(60^{\circ})}{2~sin(60^{\circ})}$
$T_b = 8.3~N$
The tension in the bottom wire is 8.3 N
We can use $T_b$ to find $T_t$.
$T_t = \frac{T_b~cos(\theta) + mg}{cos(\theta)}$
$T_t = \frac{(8.3~N)~cos(60^{\circ}) + (0.300~kg)(9.80~m/s^2)}{cos(60^{\circ})}$
$T_t = 14.2~N$
The tension in the top wire is 14.2 N