to find probability of the given problem or 
I solved this question like this but still doubt is there in my mind in the part where London is mentioned.
let us take first equally likely probability that letter could be from London or Washington
now, let $p(E)$ represent the probability that on is the only word legible then we have 
$p(E)= 0.5*p(E_1)+0.5*p(E_2)$
where $p(E_1)$ represents the probability of the only word on which is legible 
so how to calculate this $p(E_1)$ and $p(E_2)$.
Or is my approach wrong all together?!

This problem cannot be solved without further assumptions. Such an assumption is that the mail has come from London is $\frac12.$ Another assumption is that the letters in the word on the stamp delete randomly, independently and the probability that a letter gets deleted is $\frac12$ again.
What is the probability then that given WASHINGTON the remaining letters are ON? There are $10$ letters here and the only possibility is that the first $8$ letters get deleted and the last two remain. The probability is $$P(ON\mid  WASHINGTON)=\frac 1{2^{10}}=\frac1{1024}.$$
In the case of LONDON there are two possibilities and the number of letters is only $6$. So the conditional probability sought for is
$$P(ON\mid LONDON)=\frac2{2^6}=\frac1{2^5}.$$
The probability that the mail came from London provided the letters on the envelop is 
$$P(LONDON\mid ON)=\frac{P(ON\cap LONDON)}{P(ON)}=\frac{P(ON\mid LONDON)\frac12}{P(ON)}=\frac1{2^6}\frac1{P(ON)}$$
where 
$$P(ON)=P(ON\mid LONDON)\frac12+P(ON\mid WHASHINGTON)\frac12=$$
$$=\frac{\frac1{2^6}}{\frac1{2^6}+\frac1{2^{11}}}=\frac{2^5}{1+2^5}\approx 0.97.$$
While $$P(WASHINGTON\mid ON)=\frac{2^6}{2^6+2^{11}}\approx 0.03.$$
