Answer
$179\;\rm N$
Work Step by Step
First, we need to draw the force diagram of the cart, as we see below.
Now we need to apply Newton's second law in the $x$-direction which is the radial direction here.
$$\sum F_x=T\cos\theta -F_n\cos(90^\circ-\theta)=ma_r=m\dfrac{v^2}{R}$$
Thus,
$$ T\cos\theta -F_n\sin \theta =m\dfrac{v^2}{R}$$
where $v=\omega R$; plugging that we got;
$$ T\cos\theta -F_n\sin \theta =m\omega^2 R\tag 1$$
From the right triangle in the figure below, we can see that
$$\cos\theta=\dfrac{R}{L}$$
where $L$ is the length of the wire.
So, $R=L\cos\theta$
Plugging into (1);
$$ T\cos\theta -F_n\sin \theta =m\omega^2 L\cos\theta\tag 2 $$
Now we need to find the normal force,
$$\sum F_y=T\sin\theta+F_n\sin(90^\circ-\theta)- mg=ma_y=m(0)=0 $$
Thus,
$$T\sin\theta+F_n\cos \theta = mg$$
$$ F_n=\dfrac{ mg-T\sin\theta}{\cos \theta }$$
Plugging into (2);
$$ T\cos\theta -\dfrac{ mg-T\sin\theta}{\cos \theta } \sin \theta =m\omega^2 L\cos\theta $$
$$ T\cos\theta - mg\tan\theta +T\sin\theta\tan\theta =m\omega^2 L\cos\theta $$
$$ T(\cos\theta + \sin\theta\tan\theta ) =m\omega^2 L\cos\theta+mg\tan\theta $$
$$ T =\dfrac{m[\omega^2 L\cos\theta+ g\tan\theta] }{\cos\theta + \sin\theta\tan\theta} \tag3$$
Now we need to plug the given but we first need to conert the angular speed from rpm to rad per second.
$$\omega=\rm\dfrac{14\;\rm rev}{1\;min}\times \dfrac{1\;min}{60\;s}\times \dfrac{2\pi\;rad}{1\;rev}=\dfrac{7\pi}{15}$$
Plugging it and the known into (3);
$$ T =\dfrac{25\left[\left(\dfrac{7\pi}{15}\right)^2 (2)\cos20^\circ+ 9.8\tan20^\circ \right]}{\cos20^\circ + \sin20^\circ\tan20^\circ} $$
$$T\approx \color{red}{\bf 179}\;\rm N$$
