Let $$S_k = \frac{1+2+3+\ldots+k}{k}$$. If $$S_1^2 + S_2^2 + \ldots + S_{10}^2 = \frac{5}{12}A$$, then A is equal to:

The expression inside the question is $$S_k=\dfrac{1+2+3+\ldots+k}{k}.$$

First, we recall the standard formula for the sum of the first $$k$$ natural numbers:

$$1+2+3+\ldots+k=\dfrac{k(k+1)}{2}.$$

Substituting this result into the definition of $$S_k$$, we get

$$S_k=\dfrac{\dfrac{k(k+1)}{2}}{k}=\dfrac{k+1}{2}.$$

Now we need $$S_k^2$$, so we square the above expression term‐by‐term:

$$S_k^2=\left(\dfrac{k+1}{2}\right)^2=\dfrac{(k+1)^2}{4}.$$

According to the problem, we must compute $$S_1^2+S_2^2+\ldots+S_{10}^2$$. Writing each term with the squared form we just derived, we have

$$S_1^2+S_2^2+\ldots+S_{10}^2=\sum_{k=1}^{10}\dfrac{(k+1)^2}{4}=\dfrac{1}{4}\sum_{k=1}^{10}(k+1)^2.$$

To make the summation simpler, let us shift the index. Set $$j=k+1.$$ When $$k=1,\;j=2$$ and when $$k=10,\;j=11$$. Hence

$$\sum_{k=1}^{10}(k+1)^2=\sum_{j=2}^{11}j^2.$$

Instead of summing from $$2$$ to $$11$$ directly, we use the well‐known formula for the sum of squares from $$1$$ to $$n$$:

$$1^2+2^2+\ldots+n^2=\dfrac{n(n+1)(2n+1)}{6}.$$

Taking $$n=11$$ gives

$$1^2+2^2+\ldots+11^2=\dfrac{11\cdot12\cdot23}{6}.$$

We multiply step by step:

$$11\cdot12=132,$$

$$132\cdot23=3036,$$

and then divide by $$6$$:

$$\dfrac{3036}{6}=506.$$

Therefore, $$\sum_{j=1}^{11}j^2=506.$$ To remove the first term $$1^2$$ and keep only $$j=2$$ to $$11$$, we subtract $$1$$:

$$\sum_{j=2}^{11}j^2=506-1=505.$$

Substituting back into our earlier expression and remembering the factor $$\dfrac14$$ in front, we get

$$S_1^2+S_2^2+\ldots+S_{10}^2=\dfrac14\cdot505=\dfrac{505}{4}.$$

The problem tells us that this same quantity equals $$\dfrac{5}{12}A$$, so we write

$$\dfrac{5}{12}A=\dfrac{505}{4}.$$

To isolate $$A$$, we multiply both sides by $$\dfrac{12}{5}$$:

$$A=\dfrac{505}{4}\times\dfrac{12}{5}.$$

Simplifying, first divide $$12$$ by $$4$$ to obtain $$3$$:

$$A=\dfrac{505\cdot3}{5}.$$

Now divide $$505$$ by $$5$$ to get $$101$$ (or equivalently cancel the common factor $$5$$ first):

$$A=101\cdot3=303.$$

Thus the required value of $$A$$ is $$303$$.

Hence, the correct answer is Option B.