#### Answer

$1011010_{\text{two}}$

#### Work Step by Step

The place values of base two are: $...,2^6, 2^5, 2^4, 2^3, 2^2, 2^1, 1$.
The place values that are less than or equal to $90$ are $2^6, 2^5, 2^4, 2^3$, $2^2$, $2^1$, and $1$
Divide $90$ by $2^6$ or $64$ to obtain:
$90 \div 64 = 1$ remainder $26$
Divide the remainder $26$ by $2^5$ or $32$ to obtain:
$26 \div 32 = 0$ remainder $26$
Divide the remainder $26$ by $2^4$ or $16$ to obtain:
$26 \div 16 = 1$ remainder $10$
Divide the remainder $10$ by $2^3$ or $8$ to obtain:
$10 \div 8 = 1$ remainder $2$
Divide the remainder $2$ by $2^2$ or $4$ to obtain:
$2 \div 4 = 0$ remainder $2$
Divide the remainder $2$ by $2^1$ or $2$ to obtain:
$2 \div 2 = 1$ remainder $0$
Divide the remainder $0$ by $1$ to obtain:
$0 \div 1 = 0$ remainder $0$
Thus,
$90=(1 \times 2^6)+(0\times 2^5) + (1 \times 2^4) + (1 \times 2^3) + (0 \times 2^2) + (1 \times 2^1) + (0 \times 1)
\\90=1011010_{\text{two}}$