@ ashutosh 
its not a wheatstone bridge combination . so, we need to take series and parallel combination here . 
3 ohm and 4 ohm are in series , and anathor 3 ohm and 5 ohm they are also in series with each other . 
and the middle galvanometer arm are parallel with the two . 
so, frst redraw the circuit . in which 1 resistance will be 7 ohm , and its parallel to 3 ohm and again the both are parallel to 8 ohm . 
here i have applied series combination at frst R (equi ) = R1+R2 
then u have now 3 resistor in parallel , now apply . 
r ( equi ) = 1/ r1 + 1/r2 + 1/r3 
which is 
1/7 + 1/3 + 1/8 = 1/ R 
then calculate the value of R .. 
HOPE U CAN DO IT NOW 
IF ANSWER NOT MATCHING THEN LET ME KNW . I
ALL THE BEST ..
						

							
It would only be solved after applyinh KVL in loop 3-3-3 ohm resistances
and loop 3-4-5 resistances.
Or U can also tryy with KCl.
As wheat stone bridge consditions doesn't satifies here.
						

							Dear there is another method in which you can convert this into very easy circuit called delta star conversion method. In which this triangle can be converted into star like shape by applying some formulas.    It is a long process and you should to ask it to your teacher.