What is the probability that a non-leap year selected at random will contain 53 Sundays? 

We know that  

In a non leap year there are 365 days 

The number of weeks is $\frac{365}{7}= 52 weeks 1 day$ 

In a non leap year there are 52 complete weeks and 1 day which can be any day of the week i.e. Sunday, Monday, Tuesday, Wednesday, Thursday, Friday and Saturday 

Hence total number of days which can be any day is 7 

The favourable day is 1

So, favourable outcomes is1

The total outcomes is 7

$$\Rightarrow P\left(E\right)=\frac{No. of favourable outcomes}{Total outcome}$$

$$=\frac{1}{7}$$

Hence option $\left(a\right) \frac{1}{7}$  is correct