The total distance travelled by the man will be $2d$. Therefore, the time taken by the man in still water will be,

$t_0=\frac{2d}{2v}=\frac{d}{v}$

When the man is going downstream its speed will be equal to

$v_1=v_m+v_r=3v$

Therefore, the time taken to go the distance $d$ will be,

$t_1=\frac{d}{v_1}=\frac{d}{3v}$

When the man is going upstream its speed will be equal to

$v_2=v_m-v_r=v$

Therefore, the time taken to go the distance $d$ will be,

$t_2=\frac{d}{v_2}=\frac{d}{v}$

Therefore the total time taken is

$$\begin{gathered} t=t_1+t_2=\frac{4d}{3v} \\ \Rightarrow \frac{t}{t_0}=\frac{4}{3} \end{gathered}$$