Which letter cluster will replace the question mark (?) to complete the given series?
YFJW, BJEC, ?, HRUO, KVPU

To find the letter cluster that replaces the question mark (?) to complete the given series, we need to analyze the pattern of each letter position in the clusters:
YFJW, BJEC, ?, HRUO, KVPU

Let us write down the alphabetical positions (where A = 1, B = 2, ..., Z = 26) of the letters in each known cluster:

1st Letter of each cluster:
Y (25), B (2), ?, H (8), K (11)
Let us find the difference between consecutive first letters:
From Y to B: $\mathrm{25}+3=\mathrm{28}\equiv 2$ (since $\mathrm{28}-\mathrm{26}=2$, which is B). So, the pattern is $+3$.
Let's check the end of the series: From H (8) to K (11) is $8+3=11$ (K).
Applying this $+3$ pattern to B (2):
$2+3=5$, which corresponds to the letter E.
Checking if E (5) to H (8) fits: $5+3=8$ (H). This matches perfectly.

2nd Letter of each cluster:
F (6), J (10), ?, R (18), V (22)
Let us find the difference between consecutive second letters:
From F (6) to J (10): $6+4=10$ (pattern is $+4$).
From R (18) to V (22): $18+4=22$.
Applying $+4$ to J (10):
$10+4=14$, which corresponds to the letter N.
Checking if N (14) to R (18) fits: $14+4=18$ (R). This matches.

3rd Letter of each cluster:
J (10), E (5), ?, U (21), P (16)
Let us find the difference between consecutive third letters:
From J (10) to E (5): $10-5=5$ (pattern is $-5$).
From U (21) to P (16): $21-5=16$.
Applying $-5$ to E (5):
$5-5=0\equiv 26$, which corresponds to the letter Z.
Checking if Z (26) to U (21) fits: $26-5=21$ (U). This matches.

4th Letter of each cluster:
W (23), C (3), ?, O (15), U (21)
Let us find the difference between consecutive fourth letters:
From W (23) to C (3): $23+6=29\equiv 3$ (since $29-26=3$, which is C). So, the pattern is $+6$.
From O (15) to U (21): $15+6=21$.
Applying $+6$ to C (3):
$3+6=9$, which corresponds to the letter I.
Checking if I (9) to O (15) fits: $9+6=15$ (O). This matches.

Combining all the derived letters, we get ENZI.

Therefore, the correct option is ENZI.