Answer is C.
$\text{ Ordering: }\\ \text{ First Multiply } M_2 \times M_3. \text{ This requires 100*20*5 multiplications.}$ $\text{ Then Multiply } M_1 \times (M_2 \times M_3). \text{ This requires 10*100*5 multiplications. }$ $\text{ Then Multiply } (M_1 \times (M_2 \times M_3)) \times M_4. \text{ This requires 10*5*8 multiplications. }$ $\text{ Total 19000 Multiplications.}$
Brute Force approach  anyone can do.
No. of possible ordering for 4 matrices is $C_3$ where $C_3$ is the $3^{rd}$ Catalan number and given by $n=3$ in $\frac{1}{n+1} {}^{2n}C_n = 5.$
So, here we have
 $(M_1 \times M_2) \times (M_3 \times M_4)$
 $(M_1 \times (M_2 \times M_3)) \times M_4$
 $((M_1 \times M_2) \times M_3) \times M_4$
 $M_1 \times (M_2 \times (M_3 \times M_4))$
 $M_1 \times ((M_2 \times M_3) \times M_4))$
Each of these would give no. of multiplications required as
 $pqr + rst + prt $
 $qrs + pqs + pst$
 $pqr + prs + pst$
 $rst + qrt + pqt$
 $qrs + qst + pst$
The last 2 are having $qt$ terms which are the highest terms by far and hence we can avoid them from consideration $qt = 8000$ multiplied by one other term would be larger than any value in choice. So, just find the value of first 3 terms.
 $pqr + rst + prt = 20000 + 8000 + 16000 = 44000$
 $qrs + pqs + pst = 10000 + 5000 + 4000 = 19000$  smallest value in choice, we can stop here.
 $pqr + prs + pst$
Dynamic Programming Solution (should know Matrix Chain Ordering algorithm)
Here we have a chain of length 4.
Dynamic programming solution of Matrix chain ordering has the solution
$m[i, j] = \begin{cases} 0 &\text{ if } i=j\\ \min_{i\leq k < j } m[i]k] + m[k+1][j] + p_{i1}p_{j}p_{k} &\text{ if } i < j\end{cases}$
So, we can fill the following table starting with the diagonals and moving upward diagonally. Here $k < j$ but $\geq i.$

j=1 
j=2 
j=3 
j=4 
i=1 
0 
$p_0p_1p_2 =
\\ 20000$

$\min(10000+p_0p_1p_3,
\\20000+p_0+p_2p_3)
\\=15000$

$\min(18000+ p_0p_1p_4, \\20000+8000+p_0+p_2+p_4, \\15000+p_0p_3p_4) = \\19000$ 
i=2 

0 
$p_1p_2p_3 = 10000$ 
$\min(10000 + p_2p_3p_4), \\p_1p_3p_4) = 18000$ 
i=3 


0 
$p_2p_3p_4 = 8000$ 
i=4 



0 
Our required answer is given by $m[1,4] = 19000.$