If in a code language ADVENTURES is coded as VDANEUTSER, then SYSTEMATIC would be coded as:

Let's look at the arrangement pattern of the word ADVENTURES to VDANEUTSER:
Divide the 10-letter word into 5 pairs of 2 letters each:
(AD) (VE) (NT) (UR) (ES)

Now, reverse the order of letters within each individual pair:

AD ⟹ DA

VE ⟹ EV

NT ⟹ TN

UR ⟹ RU

ES ⟹ SE
This gives: DA EV TN RU SE

Let's double-check if it's paired reversal or swap across symmetric blocks:
Notice the code is VDANEUTSER. Let's split it as:
V D A N E U T S E R
Let's check the position mapping:
Original positions:
A(1) D(2) V(3) E(4) N(5) T(6) U(7) R(8) E(9) S(10)

Code:
V(3) D(2) A(1) N(5) E(4) U(7) T(6) S(10) E(9) R(8)
Wait, let's re-verify the indices:
Positions 1-2-3 (ADV) becomes VDA (3-2-1).
Positions 4-5 (EN) becomes NE (5-4).
Positions 6-7 (TU) becomes UT (7-6).
Positions 8-9-10 (RES) becomes SER (10-9-8).

Let's look at another clean split:
Split the word into two equal halves of 5 letters each:
First half: A D V E N
Second half: T U R E S

Reverse the first half: N E V D A
Reverse the second half: S E R U T
This doesn't yield VDANEUTSER either.

Let's check alternative grouping:
Pairs rearranged:
1, 2, 3 ⟹ 3, 2, 1 (ADV ⟹ VDA)
4, 5 ⟹ 5, 4 (EN ⟹ NE)
6, 7 ⟹ 7, 6 (TU ⟹ UT)
8, 9, 10 ⟹ 10, 9, 8 (RES ⟹ SER)
Total code = VDA + NE + UT + SER = VDANEUTSER.

Let's apply this exact index pattern (3,2,1, 5,4, 7,6, 10,9,8) to SYSTEMATIC (10 letters):
1:S, 2:Y, 3:S, 4:T, 5:E, 6:M, 7:A, 8:T, 9:I, 10:C

First 3 letters (SYS) reversed ⟹ SYS (3,2,1)

Next 2 letters (TE) reversed ⟹ ET (5,4)

Next 2 letters (MA) reversed ⟹ AM (7,6)

Last 3 letters (TIC) reversed ⟹ CIT (10,9,8)

Combining them: SYS + ET + AM + CIT = SYSETAMCIT. This perfectly matches Option B.