Answer
${v} =\dfrac{v_0R} {R+v_0\mu_kt}$
Work Step by Step
First of all, we need to draw the force diagram of the block.
We can see that there are two normal forces exerted on the block, one due to the base of the track and one due to the wall of the track.
Now we need to apply Newton's second law in the $x$-direction which is the radial direction of the block at any point.
$$\sum F_x=F_{n,wall}=ma_r=m\dfrac{v^2}{R}$$
$$F_{n,wall} =m\dfrac{v^2}{R}\tag 1$$
$$\sum F_z=-f_k=ma_T$$
where $f_k=\mu_kF_n$ but what normal fore should we use?
If there was friction between the block and the base of the track, then we need to use the normal force from the base. But since the friction here is between the wall and the block, we need to use the normal force due to the wall.
Thus, the friction force here is given by $f_k=\mu_kF_{n, wall}$
So that,
$$-\mu_kF_{n,wall}=ma_T$$
Plugging from (1);
$$-\mu_k\color{red}{\bf\not} m\dfrac{v^2}{R}=\color{red}{\bf\not} ma_T$$
$$-\mu_k \dfrac{v^2}{R}= a_T$$
We know that $a_T=\dfrac{dv}{dt}$
$$-\mu_k \dfrac{v^2}{R}= \dfrac{dv}{dt}$$
Hence,
$$ \dfrac{-\mu_k }{R}dt= \dfrac{ 1}{v^2}dv$$
Integrating both sides;
$$ \int_{0}^t\dfrac{-\mu_k }{R}dt= \int_{v_0}^v v^{-2}dv$$
$$\dfrac{-\mu_k }{R}t=\dfrac{v^{-1}}{-1}\bigg|_{v_0}^v$$
$$\dfrac{-\mu_k }{R}t=\dfrac{-1}{v}\bigg|_{v_0}^v=\dfrac{-1}{v}-\dfrac{-1}{v_0}$$
$$\dfrac{-\mu_k }{R}t= \dfrac{-1}{v}+\dfrac{ 1}{v_0} $$
$$ \dfrac{ 1}{v}=\dfrac{ 1}{v_0}+\dfrac{ \mu_k }{R}t $$
$$ \dfrac{ 1}{v} =\dfrac{R+v_0\mu_kt}{v_0R} $$
Therefore,,
$$ \boxed{ {v} =\dfrac{v_0R} {R+v_0\mu_kt}}$$
