A person can row $22\frac{1}{2}\text{ km}$ an hour in still water and he finds that it takes him twice as long to row up as to row down the stream. The speed of the stream is:

Let the speed of the boat in still water be $v_b$ and the speed of the stream be $v_s$.
Given, $v_b=22\frac{1}{2}\text{ km/hr}=22.5\text{ km/hr}$.

Let the distance covered be $D$.

* Downstream speed $=v_b+v_s$
* Upstream speed $=v_b-v_s$

According to the question, time taken upstream ($T_u$) is twice the time taken downstream ($T_d$):

$$T_u=2\times T_d$$

$$\frac{D}{v_b-v_s}=2\times \frac{D}{v_b+v_s}$$

$$v_b+v_s=2(v_b-v_s)$$

$$v_b+v_s=2v_b-2v_s$$

$$3v_s=v_b$$

Substitute the value of $v_b$:

$$3v_s=22.5$$

$$v_s=\frac{22.5}{3}=7.5\text{ km/hr}=7\frac{1}{2}\text{ km/hr}$$