A first order reaction A---- products has k=6.93min-1 .if 90% of first order reaction B---- product is completed in the time taken by 50%of first order reaction (A---- products).Then the rate constant for the reaction B---- products will be.

ACC to the question , both the reactions are in first order kinetics . 
so, acc to the rate law of first order kinetics 
T= 2.303/K log [A]/B[]
and here A is the total product and B is the product left after completetion of reaction . 
as u knows the k value for the reaction , put that value here , 
so, by putting the value 
T= 2.303/ 6.93 log 100/10 (as 90 percent completed ) 
calculate the value for t from the above equation . 
and as u have mentioned that the time taken for completation of 90 percent of first reaction = the time taken for completation of 50 percent of the 2^nd reaction . 
so, we can use the value of T here , also . 
use the formula 
K= 2.303/T log 100/50 
put the value of T as u have got from first equation . 
and substitute the value here , 
u will get the k2 for the 2^nd reaction . 
HOPE IT CLEARS YOUR DOUBT 
IF NOT ABLE TO FIND THE RIGHT ANSWER BY PUTTING THE VALUE THEN LET ME KNOW .. 

ALL THE BEST ..
 

I am approving you sir but i regret that its A gives rise to products now can you solve the problem for me