\text{Let a and b be roots of the equation}{\mathrm{ x}}^2-7x+c=0\text{, and the roots of the equation }{x}^2-dx+216=\mathrm{0 }\text{are }{a}^{2}b\text{ and }b{a}^2.\text{The value of}\mathrm{ d}\text{ is \_\_\_\_.}

The correct option is 42.

Here is the step-by-step derivation to find the value of d:

Let the roots of the first quadratic equation, which is:

$x^2-7x+c=0$

be a and b. Using the relationships between roots and coefficients of a quadratic equation, we have:
1. Sum of the roots:

$a+b=7$

2. Product of the roots:

$ab=c$

Now, let us analyze the second quadratic equation:

$x^2-dx+216=0$

The roots of this equation are given as:

$a^{2}b$

and

$a{b}^2$

Using the product of the roots relation for the second equation:

$\left(a^{2}b\right)\left(a{b}^2\right)=216$

Simplifying the product:

$a^3{b}^3=216$

Which is equivalent to:

${\left(ab\right)}^3=216$

Taking the cube root on both sides:

$ab=6$

Now, we have a system of equations for a and b:
Sum: a + b = 7
Product: ab = 6

We can find the values of a and b by solving the quadratic equation:

$t^2-7t+6=0$

Factoring this equation:

$(t-6)(t-1)=0$

Thus, the roots are t = 1 and t = 6. Without loss of generality, let:

$a=6$

and

$b=1$

Now, we can compute the sum of the roots of the second equation to find d. The sum of the roots of the second equation is given by d:

$d=a^{2}b+a{b}^2$

Substitute the values of a and b into this expression:

$d=(6^2\times 1)+(6\times 1^2)$

$d=36+6$

$d=42$

Therefore, the value of d is 42.