A man has a certain number of small boxes to pack into parcels. If he packs $3,4,5$ or $6$ in a parcel, he is left with one, if he packs $7$ in a parcel, none is left over. What is the number of boxes, he may have to pack?

The correct option is D. $301$
Given, if the person packs $7$ boxes then none of the box left out.
So, the total number of boxes would be multiple of $7$.
It is given that, if he packs $3,4,5$ or $6$ boxes in a parcel he is left with one box.
So, we need find a number that is multiple of $7$ and when it divided by $3,4,5$ or $6$, it leaves remainder of $1$.
Lets find the LCM of $3,4,5$ and $6$.

Thus, L.C.M of $3,4,5$ and $6$ is $3\times 2\times 2\times 5=60$
Since, the required number leaves the remainder 1 when it is divisible by $3,4,5,6$ and no remainder when it is divisible by $7$.

Therefore the required number is of the form $60x+1$

$\Rightarrow 60x+1$ is a multiple of $7$

Now, lets check for $60x+1$ is divisible by $7$ by substituting the natural numbers sequentially,

$60\left(1\right)+1=61=(7\times 8)+3$ is not divisible by $7$.

$60\left(2\right)+1=121=(7\times 17)+2$ is not divisible by $7$.

$60\left(3\right)+1=181=(7\times 25)+6$ is not divisible by $7$.

$60\left(4\right)+1=241=(7\times 34)+3$ is not divisible by $7$.

$60\left(5\right)+1=301=(7\times 43)+0$ is divisible by $7$.

So, the least number is $301$ when divided by $3,4,5,6$ leaves a remainder $1$, but when divided by $7$, there will be no remainder.

Hence, the correct option is D. $301$