We need the number of functions $f:\{1,2,3,4\}\to\{1,2,3,4,5,6\}$ such that $f(1)+f(2)=f(3).$

Let $a=f(1),\quad b=f(2),\quad c=f(3),\quad d=f(4).$ Then:

• $a,b,c,d\in\{1,2,3,4,5,6\}$
• the condition becomes $a+b=c.$

Since $c\in\{1,2,3,4,5,6\}$, we must have $a+b\le 6.$ So we need to count ordered pairs $(a,b)$ with $a,b\in\{1,2,3,4,5,6\},\quad a+b\le 6.$

Count these ordered pairs:

• If $a=1$, then $b=1,2,3,4,5$ gives $5$ choices.
• If $a=2$, then $b=1,2,3,4$ gives $4$ choices.
• If $a=3$, then $b=1,2,3$ gives $3$ choices.
• If $a=4$, then $b=1,2$ gives $2$ choices.
• If $a=5$, then $b=1$ gives $1$ choice.
• If $a=6$, then no value of $b\ge 1$ works.

Total number of ordered pairs: $5+4+3+2+1=15.$

For each such pair $(a,b)$, the value of $c=f(3)$ is uniquely determined as $c=a+b.$

Now $f(4)=d$ is unrestricted, so it can be any of the $6$ values in the codomain.

Hence total number of functions is $15\times 6=90.$

Therefore, the correct option is $\boxed{\text{B: }90}.$

Comparison with stored correct answer:

• Stored correct answer: $\text{B}$
• Derived answer: $\text{B}$
• They agree.