Explanation:

To solve this problem, let's break down the requirements:

- We need a number $x$ such that $x\equiv 7(\mathrm{mod}12)$, $x\equiv 7(\mathrm{mod}16)$, and $x\equiv 7(\mathrm{mod}54)$. This implies that $x-7$ is a common multiple of 12, 16, and 54.

- Calculate the least common multiple (LCM) of 12, 16, and 54:

- Prime factors for 12: $2^2\times 3$

- Prime factors for 16: $2^4$

- Prime factors for 54: $2\times 3^3$

- LCM is $2^4\times 3^3=432$.

- Hence, $x=432k+7$ for some integer $k$.

- $x$ must also be divisible by 13, so solve $432k+7\equiv 0(\mathrm{mod}13)$.

Now, check possible values:

- Test small values of $k$ to find divisibility by 13:

- $k=1\to 432\times 1+7=439$ (not divisible by 13)

- $k=5\to 432\times 5+7=2167$

- Check: 2167 is divisible by 13. Sum of digits: 2+1+6+7 = 16.

- Correct Option: 3. 16