⚡ AZUCATION SHORTCUT:
LHS power is $\sum_{k=1}^{20} k^2 = \frac{20 \cdot 21 \cdot 41}{6} = 2870$.
RHS power is Sum of AP from 21 to $(20+m)$.
Average of RHS $\approx 21 + \frac{m}{2}$. Terms $= m$.
$m(21 + \frac{m-1}{2}) > 2870$. Check options for $m \approx 57$.

Step 1: Simplify the LHS
Given $a_k = 3^k$.
LHS $= (3^1)^1 \times (3^2)^2 \times \dots \times (3^{20})^{20} = 3^{1^2 + 2^2 + \dots + 20^2}$.
Using the formula $\sum n^2 = \frac{n(n+1)(2n+1)}{6}$:
Power of $3$ in LHS $= \frac{20 \times 21 \times 41}{6} = 10 \times 7 \times 41 = 2870$.

Step 2: Simplify the RHS
RHS $= 3^{21} \times 3^{22} \times \dots \times 3^{20+m} = 3^{21 + 22 + \dots + (20+m)}$.
The power is an Arithmetic Progression with $m$ terms.
Sum $= \frac{m}{2} [2(21) + (m-1)1] = \frac{m(41+m)}{2}$.

Step 3: Solve the Inequality
We need: $\frac{m(41+m)}{2} > 2870 \implies m^2 + 41m > 5740$.
$\implies m^2 + 41m - 5740 > 0$.
Testing $m = 57$: $57^2 + 41(57) = 3249 + 2337 = 5586$ (Too small).
Testing $m = 58$: $58^2 + 41(58) = 3364 + 2378 = 5742$.
Since $5742 > 5740$, the smallest natural number $m$ is 58.

*Note: Re-calculating carefully for the specific boundary defined in options.

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