Step 1: Set up the total work equations

Let 1 man's 1 day work be $M$ and 1 boy's 1 day work be $B$. Total work done by $(6\text{ men} + 8\text{ boys})$ in $10$ days is equal to the work done by $(26\text{ men} + 48\text{ boys})$ in $2$ days.

$(6M + 8B) \times 10 = (26M + 48B) \times 2$

Step 2: Simplify the equation to find the relation between men and boys

Expanding both sides of the equation:

$60M + 80B = 52M + 96B$

Step 3: Solve for the efficiency ratio

Rearranging terms to find the ratio of a man's efficiency to a boy's efficiency:

$60M - 52M = 96B - 80B \implies 8M = 16B \implies M = 2B$

Step 4: Calculate total work in terms of boys

Substituting $M = 2B$ into the first total work expression:

$\text{Total Work} = (6(2B) + 8B) \times 10 = (12B + 8B) \times 10 = 200B \text{ days}$

Step 5: Calculate time taken by 15 men and 20 boys

Converting 15 men and 20 boys entirely into boys: $15M + 20B = 15(2B) + 20B = 30B + 20B = 50B$. Time taken = $\frac{200B}{50B} = 6\text{ days}$.

$T = \frac{200}{50} = 6\text{ days}$