Let the four roots be $a, b, c,$ and $d$, so that $ab=-32$.

$\underline{\textbf{Solution 1}}$ Using Vieta's formulas, we have: \begin{align*}a+b+c+d &= 18,\\ ab+ac+ad+bc+bd+cd &= k,\\ abc+abd+acd+bcd &=-200,\\ abcd &=-1984.\\ \end{align*} From the last of these equations, we see that $cd = \frac{abcd}{ab} = \frac{-1984}{-32} = 62$. Thus, the second equation becomes $-32+ac+ad+bc+bd+62=k$, and so $ac+ad+bc+bd=k-30$. The key insight is now to factor the left-hand side as a product of two binomials: $(a+b)(c+d)=k-30$, so that we now only need to determine $a+b$ and $c+d$ rather than all four of $a,b,c,d$. Let $p=a+b$ and $q=c+d$. Plugging our known values for $ab$ and $cd$ into the third Vieta equation, $-200 = abc+abd + acd + bcd = ab(c+d) + cd(a+b)$, we have $-200 = -32(c+d) + 62(a+b) = 62p-32q$. Moreover, the first Vieta equation, $a+b+c+d=18$, gives $p+q=18$. Thus we have two linear equations in $p$ and $q$, which we solve to obtain $p=4$ and $q=14$. Therefore, we have $(\underbrace{a+b}_4)(\underbrace{c+d}_{14}) = k-30$, yielding $k=4\cdot 14+30 = \boxed{86}$.

$\underline{\textbf{Solution 2}}$ We start as before: $ab=-32$ and $cd=62$. We now observe that a and b must be the roots of a quadratic, $x^2+rx-32$, where r is a constant (secretly, r is just -(a+b)=-p from Solution #1). Similarly, c and d must be the roots of a quadratic $x^2+sx+62$. Now \begin{align*}x^4-18x^3+kx^2+200x-1984 =& (x^2+rx-32)(x^2+sx+62)\\ =& x^4+(r+s)x^3+(62-32+rs)x^2\\ &+(62s-32r)x-1984.\end{align*} Equating the coefficients of $x^3$ and $x$ with their known values, we are left with essentially the same linear equations as in Solution #1, which we solve in the same way. Then we compute the coefficient of $x^2$ and get $k=\boxed{86}.$