Solution - Base conversion

$${\left(9B6\right)}_{16}$$

Expand digits with place-value powers in the source base.

$$9\cdot {16}^2+11\cdot {16}^1+6\cdot {16}^0$$

Sum the expanded terms to obtain the base-10 value.

$$2486$$

Expand each digit by place value: $$9·{16}^2$$
+ $$11·{16}^1$$
+ $$6·{16}^0$$. This gives $$2,486$$ in base 10.

$$2486\left(base10\right)$$

Use repeated division by the target base.

$$\frac{2486}{10}$$

Record quotient and remainder for this division cycle.

$$2486=10\cdot 248+6$$

Record quotient and remainder for this division cycle.

$$248=10\cdot 24+8$$

Record quotient and remainder for this division cycle.

$$24=10\cdot 2+4$$

Record quotient and remainder for this division cycle.

$$2=10\cdot 0+2$$

Read remainders from last to first to form target-base digits.

$$2486$$

Final converted representation in the target base.

$${\left(2486\right)}_{10}$$

Repeatedly divide $$2,486$$ by $$10$$, collect remainders, and read them in reverse to get $$2,486$$.