1. Arrange the letters in alphabetical order

The word is SMALL, whose letters are:

$S, M, A, L, L$

In dictionary order, the letters are:

$A < L < M < S$

Since there are two $L$'s, we must account for repetition.

2. Find how many words come before $\text{SMALL}$

We count dictionary-wise, letter by letter.

Step 1: First letter smaller than $S$

Possible smaller letters are $A, L, M$.

(i) First letter $A$

Remaining letters: $S, M, L, L$

Number of distinct arrangements:

$\frac{4!}{2!}=12$

(ii) First letter $L$

Remaining letters: $S, M, A, L$

All distinct, so number of arrangements:

$4!=24$

(iii) First letter $M$

Remaining letters: $S, A, L, L$

Number of distinct arrangements:

$\frac{4!}{2!}=12$

So total words before those starting with $S$:

$12+24+12=48$

3. Now fix first letter as $S$

We now consider words beginning with $S$ and count those before $\text{SMALL}$.

Current pattern: $S\_\_\_\_$

Remaining letters: $M, A, L, L$

Step 2: Second letter smaller than $M$

Among remaining letters, smaller than $M$ are $A, L$.

(i) Second letter $A$

Remaining letters: $M, L, L$

Number of arrangements:

$\frac{3!}{2!}=3$

(ii) Second letter $L$

Remaining letters: $M, A, L$

All distinct, so number of arrangements:

$3!=6$

Total:

$3+6=9$

So words before $SM...$ among those starting with $S$:

$48+9=57$

4. Now fix first two letters as $SM$

Pattern: $SM\_\_\_$

Remaining letters: $A, L, L$

Step 3: Third letter smaller than $A$

There is no letter smaller than $A$.

So contribution is:

$0$

5. Now fix first three letters as $SMA$

Pattern: $SMA\_\_$

Remaining letters: $L, L$

Step 4: Fourth letter smaller than $L$

No letter smaller than $L$ among remaining letters.

Contribution:

$0$

6. Now fix first four letters as $SMAL$

Remaining letter is $L$, so no further contribution.

Thus total number of words before $\text{SMALL}$ is:

$57$

Hence the rank of $\text{SMALL}$ is:

$57+1=58$

7. Compare with stored correct answer

Derived answer is:

$58^{\text{th}}$

This matches option $D$.