1. Simplify the expressions inside the square roots

We have $\sqrt{-54}=3\sqrt{6}\,i$ so $2\sqrt{-54}=6\sqrt{6}\,i.$

Hence the given expression becomes $\sqrt{3+6\sqrt{6}\,i}-\sqrt{3-6\sqrt{6}\,i}.$

2. Write each square root in the form $a+ib$

Let $\sqrt{3+6\sqrt{6}\,i}=a+ib,$ where $a,b\in\mathbb R$.

Then $(a+ib)^2=a^2-b^2+2abi=3+6\sqrt{6}\,i.$

Comparing real and imaginary parts: $a^2-b^2=3 \qquad ...(1)$ $2ab=6\sqrt{6} \implies ab=3\sqrt{6}. \qquad ...(2)$

Also, $a^2+b^2=\sqrt{3^2+(6\sqrt6)^2}=\sqrt{9+216}=15.$

Now solve: $a^2=\frac{(a^2+b^2)+(a^2-b^2)}{2}=\frac{15+3}{2}=9 \implies a=3,$ $b^2=\frac{(a^2+b^2)-(a^2-b^2)}{2}=\frac{15-3}{2}=6 \implies b=\sqrt6,$ because $2ab>0$.

Thus, $\sqrt{3+6\sqrt6\,i}=3+i\sqrt6.$

Similarly, $\sqrt{3-6\sqrt6\,i}=3-i\sqrt6$ (using the principal square root; the other branch differs by overall sign).

3. Compute the required expression

Using principal values, $\left(3+i\sqrt6\right)-\left(3-i\sqrt6\right)=2i\sqrt6.$ So its imaginary part is $2\sqrt6.$

But square roots are two-valued in algebraic context. We may also choose $\sqrt{3+6\sqrt6\,i}=-(3+i\sqrt6), \qquad \sqrt{3-6\sqrt6\,i}=3-i\sqrt6,$ which gives $-(3+i\sqrt6)-(3-i\sqrt6)=-6,$ whose imaginary part is $0$.

Or choose $\sqrt{3+6\sqrt6\,i}=3+i\sqrt6, \qquad \sqrt{3-6\sqrt6\,i}=-(3-i\sqrt6),$ which gives $6,$ whose imaginary part is again $0$.

Or choose $\sqrt{3+6\sqrt6\,i}=-(3+i\sqrt6), \qquad \sqrt{3-6\sqrt6\,i}=-(3-i\sqrt6),$ which gives $-2i\sqrt6,$ whose imaginary part is $-2\sqrt6.$

Thus, the imaginary part can be $-2\sqrt6.$

4. Check options

• A: $-2\sqrt6$ ✅ possible
• B: $6$ ❌ imaginary part cannot be real 6 here
• C: $\sqrt6$ ❌ not obtained
• D: $-\sqrt6$ ❌ not obtained

Therefore, the correct option is A.