Given function $f(x)=ax^3+bx^2+cx+41$

Use the condition $f(1)=40$ $a+b+c+41=40$ $a+b+c=-1 \quad ...(1)$

Find the first derivative $f'(x)=3ax^2+2bx+c$ Given $f'(1)=2$, so $3a+2b+c=2 \quad ...(2)$

Find the second derivative $f''(x)=6ax+2b$ Given $f''(1)=4$, so $6a+2b=4$ $3a+b=2 \quad ...(3)$

Solve the system of equations

From (3): $b=2-3a$

Substitute into (2): $3a+2(2-3a)+c=2$ $3a+4-6a+c=2$ $-3a+c=-2$ $c=3a-2$

Now substitute $b$ and $c$ into (1): $a+(2-3a)+(3a-2)=-1$ $a-3a+3a+2-2=-1$ $a=-1$

Then, $b=2-3(-1)=5$ $c=3(-1)-2=-5$

Compute $a^2+b^2+c^2$ $a^2+b^2+c^2=(-1)^2+5^2+(-5)^2$ $=1+25+25=51$

Match with options The correct option is: $\boxed{\text{B: }51}$