The correct answer is 9.

Based on the image provided, we are given the following second-order linear ordinary differential equation:
$$x^2\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}-y=0,x\ge 1$$
with the initial conditions:
$$y(x=1)=6,{\frac{dy}{dx}|}_{x=1}=2$$

This is a homogeneous Cauchy-Euler differential equation. To find the general solution, we assume a solution of the form $y=x^r$. Substituting $y$ and its derivatives into the differential equation yields the auxiliary (characteristic) equation:
$$r(r-1)+r-1=0$$

Simplifying the characteristic equation:
$$r^2-r+r-1=0$$
$$r^2-1=0$$
Thus, the roots of the equation are:
$$r=1\text{and}r=-\mathrm{1<!--\; <mi-->1}$$

The general solution is a linear combination of the two independent solutions:
$$y\left(x\right)=C_{1}x+\frac{C_2}{x}$$

Now, we differentiate the general solution to apply the derivative initial condition:
$$\frac{dy}{dx}=C_1-\frac{C_2}{x^{\mathrm{2<!--\; <mn-->2}}}$$

Applying the first initial condition $y\left(1\right)=6$:
$$C_1\left(1\right)+\frac{C_2}{1}=6\Rightarrow C_1+C_2=6\text{(Equation 1)}$$

Applying the second initial condition ${\frac{dy}{dx}|}_{x=1}=2$:
$$C_1-\frac{C_2}{1^2}=\mathrm{2<!--\; <mn-->2}\Rightarrow C_1-C_2=2\text{(Equation 2)}$$

Adding Equation 1 and Equation 2:
$$(C_1+C_2)+(C_1-C_2)=\mathrm{6<!--\; <mn-->6}+2$$
$$2C_1=8\Rightarrow C_1=4$$

Substituting $C_1=4$ back into Equation 1:
$$4+C_2=6\Rightarrow C_2=2$$

So, the particular solution to the differential equation is:
$$y\left(x\right)=4x+\frac{2}{x}$$

Finally, we calculate the value of $y$ at $x=2$:
$$y\left(2\right)=4\left(2\right)+\frac{2}{2}$$
$$y\left(2\right)=8+1=9$$